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alexira [117]
3 years ago
12

What is 4

absmiddle" class="latex-formula"> · 2\frac{1}{5}
Mathematics
1 answer:
Alecsey [184]3 years ago
8 0

well, you already know how to convert from mixed to improper, so let's simply do the multiplication, as it's very straightforward.


\bf 4\frac{2}{7}\cdot 2\frac{1}{5}\implies \cfrac{30}{7}\cdot \cfrac{11}{5}\implies \cfrac{30}{5}\cdot \cfrac{11}{7}\implies 6\cdot \cfrac{11}{7}\implies \cfrac{66}{7}\implies 9\frac{3}{7}

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\bf L.H.S = \tt \dfrac{sec\: \theta + tan \:  \theta - 1}{tan \:  \theta - sec \:  \theta + 1}  \\  \\

:  \implies \tt \dfrac{\frac{1}{cos  \: \theta}  +  \frac{sin \:  \theta}{cos \: \theta}  - 1}{  \frac{sin \:  \theta}{cos \:  \theta} -  \frac{1}{cos \:  \theta} + 1   } \:  =   \dfrac{1 + sin \:  \theta - cos \:  \theta}{sin \: \theta + cos \:  \theta} \\  \\

: \implies \tt\dfrac{ sin \:  \theta - (cos \:  \theta - 1)}{sin \: \theta + (cos \:  \theta - 1)} \:  \times  \: \dfrac{ sin \:  \theta - (cos \:  \theta - 1)}{sin \: \theta  -  (cos \:  \theta - 1)} \\  \\

: \implies \tt\dfrac{ sin^{2}  \:  \theta  + cos^{2}  \:  \theta  + 1 - 2  \: cos \:  \theta  - 2  \: sin \:  \theta \: (cos \:  \theta - 1)}{sin^{2}  \: \theta  -  (cos \:  \theta - 1)^{2} } \\  \\

: \implies \tt\dfrac{1 + 1 - 2 \:  cos \:  \theta - 2 \: sin \:  \theta  \: cos \:  \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta - 1 + 2 \: cos \:  \theta } \\  \\

: \implies \tt\dfrac{2 - 2 \:  cos \:  \theta - 2 \: sin \:  \theta  \: cos \:  \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta  - sin^{2} \:  \theta - cos^{2}   \:  \theta  + 2 \: cos \:  \theta } \\  \\

: \implies \tt\dfrac{2 (1 - \:  cos \:  \theta )- 2 \: sin \:  \theta  (1 - \: cos \:  \theta)}{ 2 \: cos \: \theta - 2 \: cos^{2}   \:  \theta} \\  \\

: \implies \tt\dfrac{(2  +  2 \:  sin \:  \theta)  \:  \cancel{(1 -  cos\:  \theta)}}{2 \: cos \:  \theta  \:  \cancel{(1 - cos \:  \theta)}} \:  =  \:  \dfrac{1 + sin \:  \theta}{cos \: \theta}  \\  \\

: \implies\tt\dfrac{1 + sin \:  \theta}{cos \: \theta}  \:  \times  \: \dfrac{1  -  sin \:  \theta}{1 - sin \: \theta} \\  \\

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:  \implies\tt\dfrac{cos \:  \theta}{1 - sin \: \theta}  \:  = \:  \bf{ R.H.S}\\  \\

\huge\bigstar  \:\underline{\red{\sf Hence, Proved}} \:  \bigstar \\

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